Prove that if $f:D\to D$ is a one-to-one analytic mapping of the unit disc $|z|<1$ onto itself withtwo distinct fixed points, then $f(z)=z$.
(A complex number $w\in D$ is a fixed point for the map $f:D \to D$ if $f(w)=w$. )
I found an answer here but I have two questions:
(1) Would the extra condition of f being one-to-one/injective simplify @Kavi Rama Murthy's answer?
(2) For his answer and comment, why does $h(z)=z*e^{ic}$, show $h(z)=z$?
It is only needed that $f$ maps the unit disk $D$ into itself and has two fixed points $z_1, z_2$. The fact that $f$ is injective does not simplify the solution.
The same proof works also if $D \subsetneq \Bbb C$ is any simply connected domain, because of the Riemann mapping theorem.
$h=g\circ f \circ g^{-1}$ îs constructed such that is maps the unit disk into itself and has two fixed points $g(z_1) = 0$ and $g(z_2) = w^* \ne 0$. The Schwarz Lemma states that $|h(z)| \le |z|$, and then that $h(z) = e^{ic}z$ (for some $c \in \Bbb R$) because $|w^*| = |g(w^*)|$. Finally, $w^* = g(w^*) = e^{ic} w^*$ shows hat $e^{ic} = 1$, i.e. that $h$ is the identity function.
